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Forum Geometricorum Volume 4 (2004) 117–124.

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FORUM GEOM ISSN 1534-1178

Three Pairs of Congruent Circles in a Circle Li C. Tien

Abstract. Consider a closed chain of three pairs of congruent circles of radii a, b, c. The circle tangent internally to each of the 6 circles has radius R = a + b + c if and only if there is a pair of congruent circles whose centers are on a diameter of the enclosing circle. Non-neighboring circles in the chain may overlap. Conditions for nonoverlapping are established. There can be a “central circle” tangent to four of the circles in the chain.

1. Introduction Consider a closed chain of three pairs of congruent circles of radii a, b, c, as shown in Figure 1. Each of the circles is tangent internally to the enclosing circle (O) of radius R and tangent externally to its two neighboring circles.

C

B Q P

A

O C

Figure 1A: (abcacb)

B

C

A

B

O

A

B

A

C

Figure 1B: (abcabc)

The essentially distinct arrangements, depending on the number of pairs of congruent neighboring circles, are (A): (aabcbc) (B): (aacbbc) (C): (aabbcc) (D): (aaaabb) (E): (abcabc), (abcacb) (F): (aabaab), (aaabab) (G): (aaaaaa) Figures 1A and 1B illustrate the pattern (E). Patterns (D) and (F) have c = a. In pattern (G), b = c = a. Publication Date: July 30, 2004. Communicating Editor: Paul Yiu.

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According to [1, 3], in 1877 Sakuma proved R = a + b + c for patterns (E). Hiroshi Okumura [1] published a much simpler proof. Unaware of this, Tien [4] rediscovered the theorem in 1995 and published a similar, simple proof. It is easy to see by symmetry that in each of the patterns (E), (F), (G), there is a pair of congruent circles with centers on a diameter of the enclosing circle. Let us call such a pair a diametral pair. Here is a stronger theorem: Theorem 1. In a closed chain of three pairs of congruent circles of radii a, b, c tangent internally to a circle of radius R, R = a + b + c if and only if the closed chain contains a diametral pair of circles. In Figure 1, two non-neighboring circles intersect. The proof for R = a + b + c does not forbid such an intersection. Sections 4 and 5 are about avoiding intersecting circles and about adding a “central” circle. 2. Preliminaries In Figure 1, the enclosing circle (O) of radius R centers at O and the circles (A), (B), (C) of radii a, b, c, center at A, B, C, respectively. The circles (A ), (B  ), (C  ) are also of radii a, b, c respectively. Suppose two circles (A) and (B) of radii a and b are tangent externally each other, and each tangent internally to a circle O(R). We denote the magnitude of angle AOB by θab . See Figure 2A. This clearly depends on R. If a < R2 , then we can also speak of θaa . Note that the center O is outside each circle of radius a. R θaa a 2 , sin 2 = R−a . (See Figure 2A). (R−b)2 +(R−c)2 −(b+c)2 . (See Figure 2B). 2(R−b)(R−c)

Lemma 2. (a) If a < (b) cos θbc =

b+c

C

B

A R−a θab

R−c

a

O

R−b O

B

Figure 2A

Figure 2B



Proof. These are clear from Figures 2A and 2B. Lemma 3. If a and b are unequal and each <

R 2,

then θaa + θbb > 2θab .

Three pairs of congruent circles in a circle

119

Proof. In Figure 1A, consider angle AOP , where P is a point on the circle (A). The angle AOP is maximum when line OP is tangent to the circle (A). This maximum is θaa 2 ≥ ∠AOQ, where Q is the point of tangency of (A) and (B). θbb  Similarly, 2 ≥ ∠BOQ, and the result follows. Corollary 4. If a, b, c are not the same, then θaa + θbb + θcc > θab + θbc + θca . Proof. Write θaa + θbb + θcc =

θaa + θbb θbb + θcc θcc + θaa + + 2 2 2

and apply Lemma 3.



3. Proof of Theorem 1 Sakuma, Okumura [1] and Tien [4] have proved the sufficiency part of the theorem. We need only the necessity part. This means showing that for distinct a, b, c in patterns (A) through (D) which do not have a diametral pair of circles, the assumption of R = a + b + c causes contradictions. In patterns (E) with a pair of diametral circles and R = a + b + c, the sum of the angles around the center O of the enclosing circle is 2(θab + θbc + θca ) = 2π, that is, θab + θbc + θca = π. Pattern (A): (aabcbc). The sum of the angles around O is θaa + θab + θbc + θcb + θbc + θca =θab + θbc + θca + (θaa + 2θbc ) =π + (θaa + 2θbc ). This is 2π if and only if (θaa + 2θbc ) = π, or π2 − θaa 2 = θbc . The cosines of these angles, Lemma 2 and the assumption R = a + b + c lead to a2 + ab + ac − bc a = , b+c (a + b)(a + c) which gives (a − b)(a − c)(a + b + c) = 0, an impossibility, if a, b, c are distinct. Pattern (B): (aacbbc). If a > R2 or b > R2 , then the neighboring tangent circles of radii a or b, respectively, cannot fit inside the enclosing circle of radius R = a + b + c. For this equation to hold, it must be that a ≤ R2 and b ≤ R2 . Then, O is outside A(a) and B(b). The sum of the angles around O exceeds 2π, by Lemma 3: θaa + θac + θcb + θbb + θbc + θca =(θaa + θbb ) + 2(θbc + θca ) >2(θab + θbc + θca ) =2π.

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Patterns (C) and (D): (aabbcc) and (aaaabb). For R = a + b + c to hold, O must be outside A(a), B(b), C(c). Again, the sum of the angles around O exceeds 2π. For pattern (C), θaa + θab + θbb + θbc + θcc + θca =(θaa + θbb + θcc ) + (θab + θbc + θca) >(θab + θbc + θca) + (θab + θbc + θca) =2π. Here, the inequality follows from Corollary 4 for a, b, c, not all the same. For pattern (D) with c = a, the inequality remains true. This completes the proof of Theorem 1. Remark. A narrower version of Theorem 1 treats a, b, c as variables, instead of any particular lengths. The proof for this version is simple. We see that when no pair of the enclosed circles is diametral, at least one pair has its two circles next to each other. Let these two be point circles and let the other four circles be of the same radius. Then the six circles become three equal tangent circles tangentially enclosed in a circle. In this special case R = a + b + c = 0 + a + a is false. Then, a, b, c cannot be variables. 4. Nonoverlapping arrangements Patterns (A) through (G) are adaptable to hands-on activities of trying to fit chains of three pairs of congruent circles into an enclosing circle of a fixed radius R. Most of the essential patterns have inessential variations. Assuming a ≤ b ≤ c, patterns (E) have four variations: E1 E2 E3 E4

: : : :

(abcabc) (cabcba) (abcacb) (bcabac)

For hands-on activities, it is desirable to find the conditions for the enclosed circles in patterns (E) not to overlap. We find the bounds of the ratio Ra in these patterns. 4.1. Patterns E1 and E2 . The largest circles (C) and C ) are diametral. For a nonoverlapping arrangement, Clearly, a ≤ 13 R and c ≤ 12 R. In Figure 3, a circle of radius b is tangent externally to the two diametral circles of radii c, and internally to the enclosing circle of radius R. From (b + c)2 = (R − b )2 + (R − c)2 , we have b = R(R−c) R+c . It follows that in a nonoverlapping patterns E1 and E2 , with 1 1 3 R ≤ c ≤ 2 R, we have b + c ≤ b + c =

5 R2 + c2 ≤ R. R+c 6

Three pairs of congruent circles in a circle

121

b

c

B

C

c

O

C

B

Figure 3

A

A

Figure 4

From this, a ≥ 16 R. Figure 4 shows a nonoverlapping arrangement with a = 16 R, b = 13 R, c = 12 R. It is clear that for every a satisfying 16 R ≤ a ≤ 13 R, there are nonoverlapping patterns E1 and E2 (with a ≤ b ≤ c). 4.2. Patterns E3 and E4 . In these cases the largest circles (C) and (C ) are not diametral. Lemma 5. If three circles of radii x, z, z are tangent externally to each other, and are each tangent internally to a circle of radius R, then 4Rx(R − x) . z= (R + x)2

Z X

O Z

Figure 5

Proof. By the Descartes circle theorem [2], we have     1 2 2 1 2 1 1 + + , = − + + 2 R2 x2 z 2 R x z from which the result follows.



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Theorem 6. For a given R, a nonoverlapping arrangement of pattern E3 (abcacb) or E4 (bcabac) with a ≤ b ≤ c and a + b + c = R exists if γR ≤ a ≤ 13 R, where   √ √ 3 3 1 + 19 + 12 87 + 19 − 12 87 ≈ 0.25805587 · · · . γ= 6 Proof. For b = a and the largest c = R − 2a for a nonoverlapping arrangement E3 (abcacb), Lemma 5 gives f ( Ra )R3 4Ra(R − a) − (R − 2a) = = 0, (R + a)2 (R + a)2 where f (x) = 2x3 − x2 + 4x − 1. It has a unique real root γ given above.

C

B

A

A

C

B

Figure 6

Figure 6 shows a nonoverlapping arrangement E3 with a = b = γR, and c = (1 − 2γ)R. For γR ≤ a ≤ 13 R, from the figure we see that (C) and (C ) and the other circles cannot overlap in arrangements of patterns E3 (abcacb) and  E4 (bcabac). Corollary 7. The sufficient condition γR ≤ a ≤ 13 R also applies to patterns E1 and E2 . Outside the range γR ≤ a ≤ R3 , patterns E3 (abcacb) and E4 (bcabac) still can have nonoverlapping circles. Both of the patterns involve Figure 5 and z = 4Rx(R−x) , with z = c, x = a or b, and a ≤ b ≤ c. (R+x)2 √ The equation gives the smallest x √ = a = (3 − 2 2)R ≈ 0.1715 · · · R corresponding to the largest b = c = ( 2 − 1)R ≈ 0.4142 · · · R and the largest x = b = R3 corresponding to the largest c = R2 . Thus, the nonoverlapping condi√ . tions are (3 − 2 2)R ≤ x ≤ R3 and c ≤ 4Rx(R−x) R+x)2 For x ≥ R3 , circles (Z) and (Z  ) overlap with (X  ), which is diametral with  (X). Now Figure 3 and the associated b = R(R−c) R+c are relevant. With b replaced

by c and c by b, the equation becomes c = R(R−b) R+b . By this equation, when b varies

Three pairs of congruent circles in a circle

123

√ √ from R3 to ( 2−1)R, c ≥ b varies from R2 to ( 2−1)R. Thus, the nonoverlapping √ √ . The case of b > ( 2 − 1)R conditions are R3 ≤ b ≤ ( 2 − 1)R and c ≤ R(R−b) R+b makes b > c and the largest pair of circles diametral, already covered in §4.1. 5. The central circle and avoiding intersecting circles Obviously, pattern (G) (aaaaaa) admits a “central” circle tangent to all 6 circles of radii a. In patterns (F) (aabaab), (aaabab), we can add a central circle tangent to the four circles of radius a. Figure 7 shows the less obvious central circle for (abcacb) of pattern (E).

C

C

B

A

O A C

A

A B

B O

C

Figure 7

A

A B

Figure 8

Theorem 8. Consider a closed chain of pattern (abcacb). There is a “central” circle of radius a tangent to the four circles of radii b and c. This circle does not overlap with the circle A(a) if a≤ where b ≤ c.

b(b + c) , 2c

Proof. In Figure 7, the pattern of the chain tells that R = a + b + c. The central circle centered at A has radius a is tangent to B(b), B (b), C(c), C  (c) because triangles A BC and OBC are mirror images of each other. When b < c, A (a) is closer to A(a) than A (a). If A (a) and A(a) are tangent to each other, then AB 2 − a2 = OB 2 − (OA − a)2 . Now, AB = a + b and OB = a + c, OA = b + c. b(b+c)  This simplifies into a = b(b+c) 2c . If a < 2c , the circles A(a) and A (a) are separate.  Figure 8 shows an arrangement (abcacb) with a central circle touching 5 inner circles except (A ). References [1] H. Okumura, Circle patterns arising from a seven-circle problem, Crux Math., 21 (1995) 213– 217. [2] D. Pedoe, On a theorem of geometry, Amer. Math. Monthly, 74 (1967) 627–640.

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[3] J. F. Rigby, Circle problems arising from Wasan, Symmetry: Culture and Science, 8 (1997) 68– 73. [4] L. C. Tien, Constant-sum figures, Math. Intelligencer, 23 (2001) no. 2, 15–16. Li C. Tien: 4412 Huron Drive, Midland, Michigan, 48642-3503, USA E-mail address: [email protected]

Three Pairs of Congruent Circles in a Circle

Jul 30, 2004 - Abstract. Consider a closed chain of three pairs of congruent circles of radii a, b, c. The circle tangent internally to each of the 6 circles has radius R = a + b + c if and only if there is a pair of congruent circles whose centers are on a diameter of the enclosing circle. Non-neighboring circles in the chain may.

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